![[sol-5.pdf#page=28]] --- $\textbf{Exercise 14.}$ Define $T: \mathbb{R}^3 \to \mathbb{R}^2$ by $\begin{equation*} T(x, y, z) = (4x + 5y + 6z, 7x + 8y + 9z). \end{equation*}$ Suppose $\varphi_1, \varphi_2$ denotes the dual basis of the standard basis of $\mathbb{R}^2$ and $\psi_1, \psi_2, \psi_3$ denotes the dual basis of the standard basis of $\mathbb{R}^3$. (a) Describe the linear functionals $T^*(\varphi_1)$ and $T^*(\varphi_2)$. (b) Write $T^*(\varphi_1)$ and $T^*(\varphi_2)$ as linear combinations of $\psi_1, \psi_2, \psi_3$. $\textbf{Solution 14.}$ (a) Note that $\varphi_1(a, b) = a$ and $\varphi_2(a, b) = b$ for all $(a, b) \in \mathbb{R}^2$. Recall that $T^*(\varphi_1) = \varphi_1 \circ T$ and $T^*(\varphi_2) = \varphi_2 \circ T$. Thus $\begin{equation*} (T^*(\varphi_1))(x, y, z) = 4x + 5y + 6z \end{equation*}$ and $\begin{equation*} (T^*(\varphi_2))(x, y, z) = 7x + 8y + 9z. \end{equation*}$ (b) We have $\psi_1(x, y, z) = x$, $\psi_2(x, y, z) = y$, and $\psi_3(x, y, z) = z$. Using (a), we thus see that $\begin{align*} T^*(\varphi_1) &= 4\psi_1 + 5\psi_2 + 6\psi_3 \\ T^*(\varphi_2) &= 7\psi_1 + 8\psi_2 + 9\psi_3. \end{align*}$ --- $\textbf{Exercise 14.}$ Define $T: \mathbb{R}^3 \to \mathbb{R}^2$ by $\begin{equation*} T(x, y, z) = (4x + 5y + 6z, 7x + 8y + 9z). \end{equation*}$ Suppose $\varphi_1, \varphi_2$ denotes the dual basis of the standard basis of $\mathbb{R}^2$ and $\psi_1, \psi_2, \psi_3$ denotes the dual basis of the standard basis of $\mathbb{R}^3$. (a) Describe the linear functionals $T^*(\varphi_1)$ and $T^*(\varphi_2)$. (b) Write $T^*(\varphi_1)$ and $T^*(\varphi_2)$ as linear combinations of $\psi_1, \psi_2, \psi_3$. $\textbf{Solution 14.}$ (a) Note that $\varphi_1(a, b) = a$ and $\varphi_2(a, b) = b$ for all $(a, b) \in \mathbb{R}^2$. Recall that $T^*(\varphi_1) = \varphi_1 \circ T$ and $T^*(\varphi_2) = \varphi_2 \circ T$. Thus $\begin{equation*} (T^*(\varphi_1))(x, y, z) = 4x + 5y + 6z \end{equation*}$ and $\begin{equation*} (T^*(\varphi_2))(x, y, z) = 7x + 8y + 9z. \end{equation*}$ (b) We have $\psi_1(x, y, z) = x$, $\psi_2(x, y, z) = y$, and $\psi_3(x, y, z) = z$. Using (a), we thus see that $\begin{align*} T^*(\varphi_1) &= 4\psi_1 + 5\psi_2 + 6\psi_3 \\ T^*(\varphi_2) &= 7\psi_1 + 8\psi_2 + 9\psi_3. \end{align*}$ $\textit{Commentary:}$ This exercise is a concrete example of how the dual map acts on specific linear functionals. Given a linear map $T: \mathbb{R}^3 \to \mathbb{R}^2$, the dual map $T^*$ sends linear functionals on $\mathbb{R}^2$ to linear functionals on $\mathbb{R}^3$. Part (a) asks to describe these functionals explicitly, which is done by composing the original functionals with $T$. Part (b) then asks to express these functionals in terms of the dual basis of $\mathbb{R}^3$, which is a straightforward calculation using the results of part (a). This exercise helps to develop a concrete understanding of the dual map and its relationship to bases and coordinate representations. $\textit{Example:}$ Let $\varphi = (1, 2) \in (\mathbb{R}^2)^*$, so $\varphi(a, b) = a + 2b$. Then $\begin{align*} T^*(\varphi)(x, y, z) &= \varphi(T(x, y, z)) \\ &= \varphi(4x + 5y + 6z, 7x + 8y + 9z) \\ &= (4x + 5y + 6z) + 2(7x + 8y + 9z) \\ &= 18x + 21y + 24z \\ &= 18\psi_1(x, y, z) + 21\psi_2(x, y, z) + 24\psi_3(x, y, z). \end{align*}$ Thus $T^*(\varphi) = 18\psi_1 + 21\psi_2 + 24\psi_3$.